[math-fun] how not to handle equations in lectures
When I give a lecture, if I write an equation, I explain all the meanings of the letters inside it, and what it says. If Nima Arkani-Hamed flashes an equation on screen, he never (?) defines the letters inside it, never explains what it says, and just sort of waves his hand at it, kind of as though pointing out a passing airplane, then moves on. Also, he never writes a theorem statement, nor even an explicit "CLAIM: xxxx." I would say "never" except maybe sometimes he does give a very partial stab at it for a small subset of equations, so perhaps we should give him partial credit on a few equations. I don't understand how one can adopt this lecture style, except as a joke. If other physicists imitate him in this respect (since he's a great $3M prize winner) it would be a disaster. just my opinion...
Who awarded him the $3M prize, and for what? Did he happen to win a lottery? -- Gene
________________________________ From: Warren D Smith <warren.wds@gmail.com> To: math-fun@mailman.xmission.com Sent: Thursday, September 19, 2013 4:46 PM Subject: [math-fun] how not to handle equations in lectures
When I give a lecture, if I write an equation, I explain all the meanings of the letters inside it, and what it says.
If Nima Arkani-Hamed flashes an equation on screen, he never (?) defines the letters inside it, never explains what it says, and just sort of waves his hand at it, kind of as though pointing out a passing airplane, then moves on.
Also, he never writes a theorem statement, nor even an explicit "CLAIM: xxxx."
I would say "never" except maybe sometimes he does give a very partial stab at it for a small subset of equations, so perhaps we should give him partial credit on a few equations.
I don't understand how one can adopt this lecture style, except as a joke. If other physicists imitate him in this respect (since he's a great $3M prize winner) it would be a disaster.
just my opinion...
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http://www.nytimes.com/2012/07/31/science/9-scientists-win-yuri-milners-fund... On Sep 19, 2013, at 8:23 PM, Eugene Salamin <gene_salamin@yahoo.com> wrote:
Who awarded him the $3M prize, and for what?
Consider the unit n-cube and a largest regular n-simplex -- by volume -- that can fit inside it. What is V(n), the n-dimensional volume of that simplex, as a function of n? I think the first three of these are V(1) = 1, V(2) = 2sqrt(3) - 3, and V(3) = 1/3. If this is too hard to calculate in general, maybe there is an asymptotic formula for V(n) ? --Dan
On 9/20/13, Dan Asimov <dasimov@earthlink.net> wrote:
Consider the unit n-cube and a largest regular n-simplex -- by volume -- that can fit inside it.
What is V(n), the n-dimensional volume of that simplex, as a function of n?
I think the first three of these are V(1) = 1, V(2) = 2sqrt(3) - 3, and V(3) = 1/3.
If this is too hard to calculate in general, maybe there is an asymptotic formula for V(n) ?
--Dan
Warren Smith: << If you go down up to 3 dimensions to get 3 mod 4, use Hadamard to inscribe regular simplex in cube-face (of the reduced dimension) then build back up, then you will not lose out too horribly (although this will not be optimal). >> So a casual glance suggests that we might get an estimate for the answer in 4-space by fitting the base of a regular simplex of edge sqrt2 into one cube facet of the unit tesseract, giving [0, 0, 0, 0], [0, 0, 1, 1], [0, 1, 0, 1], [0, 1, 1, 0], [s, 1/2, 1/2, 1/2] where s = sqrt(5/4) . However this doesn't apparently work very well: for one thing, the altitude s > 1 , so the simplex must be scaled down to edge sqrt(8/5) to fit inside the tesseract. For another, the configuration turns out to be a local maximum: any near-identity similarity which retains the simplex within the tesseract also shrinks it. The maximal regular simplex turns out instead to have vertex coordinates [1, 0, 0, 0], [0, 1, 0, 0], [0, 0, 1, 0], [0, 0, 0, 1], [t/2, t/2, t/2, t/2] where t = (sqrt5 + 1)/2 is the "golden section" number tau, phi, etc. Which also has edge sqrt2 --- despite everything, Warren's estimate gave the correct answer exactly! The corresponding content V(4) = rt5/24 ~ 0.093169 . Fred Lunnon
Heady with success in 4-space, I embarked enthusiastically upon a general program for n-space --- only to recall, yet again, that it is better to travel hopefully than to arrive. The 5-space case is truly horrible: there are evidently multiple local maxima all with side approximately 1.5, corresponding to simplex vertices lying on various combinations of boundary elements of the hypercube. While one simplex vertex always nestles securely in a hypercube vertex, just deciding whether other simplex vertices actually meet a hypercube line, plane etc. is in practice difficult because of ill-conditioning. My most persuasive candidate so far in 20 attempts, by reason of symmetry, has also the second largest known side = 1.510898923478 ; its vertices are [1, 1, 1, a, b, c], [1, c, b, 1, a, 1], [1, a, c, b, 1, 1], [1, 1, a, c, 1, b], [1, b, 1, 1, c, a], [1, 0, 0, 0, 0, 0], where a := 0.003086685229, b := 0.089689781542, c := 0.524177233794; though these constants don't appear to be anything simply algebraic. But alas for elegance --- a much gnarlier specimen has just turned up with side = 1.511084, making it fairly certain that the last word remains to be spoken ... Fred Lunnon On 9/21/13, Fred Lunnon <fred.lunnon@gmail.com> wrote:
On 9/20/13, Dan Asimov <dasimov@earthlink.net> wrote:
Consider the unit n-cube and a largest regular n-simplex -- by volume -- that can fit inside it.
What is V(n), the n-dimensional volume of that simplex, as a function of n?
I think the first three of these are V(1) = 1, V(2) = 2sqrt(3) - 3, and V(3) = 1/3.
If this is too hard to calculate in general, maybe there is an asymptotic formula for V(n) ?
--Dan
Warren Smith: << If you go down up to 3 dimensions to get 3 mod 4, use Hadamard to inscribe regular simplex in cube-face (of the reduced dimension) then build back up, then you will not lose out too horribly (although this will not be optimal). >>
So a casual glance suggests that we might get an estimate for the answer in 4-space by fitting the base of a regular simplex of edge sqrt2 into one cube facet of the unit tesseract, giving [0, 0, 0, 0], [0, 0, 1, 1], [0, 1, 0, 1], [0, 1, 1, 0], [s, 1/2, 1/2, 1/2] where s = sqrt(5/4) .
However this doesn't apparently work very well: for one thing, the altitude s > 1 , so the simplex must be scaled down to edge sqrt(8/5) to fit inside the tesseract. For another, the configuration turns out to be a local maximum: any near-identity similarity which retains the simplex within the tesseract also shrinks it.
The maximal regular simplex turns out instead to have vertex coordinates [1, 0, 0, 0], [0, 1, 0, 0], [0, 0, 1, 0], [0, 0, 0, 1], [t/2, t/2, t/2, t/2] where t = (sqrt5 + 1)/2 is the "golden section" number tau, phi, etc. Which also has edge sqrt2 --- despite everything, Warren's estimate gave the correct answer exactly!
The corresponding content V(4) = rt5/24 ~ 0.093169 .
Fred Lunnon
At this juncture, the intelligent thing to do is abandon my gruesomely slowly-convergent search program, and instead solve for a,b,c maximising the side length of an equilateral simplex having this nice symmetric form. Lagrange finds only a single interior minimum; escaping to the nearby boundary by setting a = 0 , Lagrange locates a single maximum. The upshot has side = 1.511173070871355 with vertices [0, 0, 0, 0, 0], [1, 1, 0, c, b], [1, 0, c, b, 1], [0, c, b, 1, 1], [c, b, 1, 1, 0], [b, 1, 1, 0, c] where b = 0.09303505049476144, c = 0.5243934872843100 are quartic irrationals such that b^4 - 6*b^3 + 14*b^2 - 12*b + 1 = 0 and c = b^3 - 4*b^2 + 6*b . The 5-cyclic symmetry of the complete configuration is obvious above; all but one simplex vertex lie on hypercube (plane) faces. This must almost certainly be the overall maximal possible regular simplex inscribable in a unit hypercube in 5-space. Fred Lunnon On 9/23/13, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Heady with success in 4-space, I embarked enthusiastically upon a general program for n-space --- only to recall, yet again, that it is better to travel hopefully than to arrive.
The 5-space case is truly horrible: there are evidently multiple local maxima all with side approximately 1.5, corresponding to simplex vertices lying on various combinations of boundary elements of the hypercube. While one simplex vertex always nestles securely in a hypercube vertex, just deciding whether other simplex vertices actually meet a hypercube line, plane etc. is in practice difficult because of ill-conditioning.
My most persuasive candidate so far in 20 attempts, by reason of symmetry, has also the second largest known side = 1.510898923478 ; its vertices are [1, 1, 1, a, b, c], [1, c, b, 1, a, 1], [1, a, c, b, 1, 1], [1, 1, a, c, 1, b], [1, b, 1, 1, c, a], [1, 0, 0, 0, 0, 0], where a := 0.003086685229, b := 0.089689781542, c := 0.524177233794; though these constants don't appear to be anything simply algebraic.
But alas for elegance --- a much gnarlier specimen has just turned up with side = 1.511084, making it fairly certain that the last word remains to be spoken ...
Fred Lunnon
On 9/21/13, Fred Lunnon <fred.lunnon@gmail.com> wrote:
On 9/20/13, Dan Asimov <dasimov@earthlink.net> wrote:
Consider the unit n-cube and a largest regular n-simplex -- by volume -- that can fit inside it.
What is V(n), the n-dimensional volume of that simplex, as a function of n?
I think the first three of these are V(1) = 1, V(2) = 2sqrt(3) - 3, and V(3) = 1/3.
If this is too hard to calculate in general, maybe there is an asymptotic formula for V(n) ?
--Dan
Warren Smith: << If you go down up to 3 dimensions to get 3 mod 4, use Hadamard to inscribe regular simplex in cube-face (of the reduced dimension) then build back up, then you will not lose out too horribly (although this will not be optimal). >>
So a casual glance suggests that we might get an estimate for the answer in 4-space by fitting the base of a regular simplex of edge sqrt2 into one cube facet of the unit tesseract, giving [0, 0, 0, 0], [0, 0, 1, 1], [0, 1, 0, 1], [0, 1, 1, 0], [s, 1/2, 1/2, 1/2] where s = sqrt(5/4) .
However this doesn't apparently work very well: for one thing, the altitude s > 1 , so the simplex must be scaled down to edge sqrt(8/5) to fit inside the tesseract. For another, the configuration turns out to be a local maximum: any near-identity similarity which retains the simplex within the tesseract also shrinks it.
The maximal regular simplex turns out instead to have vertex coordinates [1, 0, 0, 0], [0, 1, 0, 0], [0, 0, 1, 0], [0, 0, 0, 1], [t/2, t/2, t/2, t/2] where t = (sqrt5 + 1)/2 is the "golden section" number tau, phi, etc. Which also has edge sqrt2 --- despite everything, Warren's estimate gave the correct answer exactly!
The corresponding content V(4) = rt5/24 ~ 0.093169 .
Fred Lunnon
Hurtling unstoppably on whither angles fear to tread, a quick blunder around in 6-space led surprisingly quickly to critical (maximal?) side and content sqrt(5/2) = 1.581138830084190, 0.007177059763087539 with vertices [b,c,b,c,b,c], [1,a,0,1,1,1], [0,1,1,1,1,a], [1,1,1,a,0,1], [1,1,a,0,1,0], [a,0,1,0,1,1], [1,0,1,1,a,0]] where b = 0.08856217223385236, c = 0.3628540574112841 satisfy b^2 - (3/2)b + 1/8 = 0, c = (b + 1)/3 . There is 3-cyclic symmetry, and simplex vertices lie on hypercube edges, except for a single one in the interior. As Warren pointed out, in 7-space we have the obviously maximal Hadamard simplex with side = 2 , sharing all vertices with the hypercube. Fred Lunnon On 9/23/13, Fred Lunnon <fred.lunnon@gmail.com> wrote:
At this juncture, the intelligent thing to do is abandon my gruesomely slowly-convergent search program, and instead solve for a,b,c maximising the side length of an equilateral simplex having this nice symmetric form. Lagrange finds only a single interior minimum; escaping to the nearby boundary by setting a = 0 , Lagrange locates a single maximum.
The upshot has
side = 1.511173070871355
with vertices
[0, 0, 0, 0, 0], [1, 1, 0, c, b], [1, 0, c, b, 1], [0, c, b, 1, 1], [c, b, 1, 1, 0], [b, 1, 1, 0, c]
where
b = 0.09303505049476144, c = 0.5243934872843100
are quartic irrationals such that b^4 - 6*b^3 + 14*b^2 - 12*b + 1 = 0 and c = b^3 - 4*b^2 + 6*b .
The 5-cyclic symmetry of the complete configuration is obvious above; all but one simplex vertex lie on hypercube (plane) faces.
This must almost certainly be the overall maximal possible regular simplex inscribable in a unit hypercube in 5-space.
Fred Lunnon
On 9/23/13, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Heady with success in 4-space, I embarked enthusiastically upon a general program for n-space --- only to recall, yet again, that it is better to travel hopefully than to arrive.
The 5-space case is truly horrible: there are evidently multiple local maxima all with side approximately 1.5, corresponding to simplex vertices lying on various combinations of boundary elements of the hypercube. While one simplex vertex always nestles securely in a hypercube vertex, just deciding whether other simplex vertices actually meet a hypercube line, plane etc. is in practice difficult because of ill-conditioning.
My most persuasive candidate so far in 20 attempts, by reason of symmetry, has also the second largest known side = 1.510898923478 ; its vertices are [1, 1, 1, a, b, c], [1, c, b, 1, a, 1], [1, a, c, b, 1, 1], [1, 1, a, c, 1, b], [1, b, 1, 1, c, a], [1, 0, 0, 0, 0, 0], where a := 0.003086685229, b := 0.089689781542, c := 0.524177233794; though these constants don't appear to be anything simply algebraic.
But alas for elegance --- a much gnarlier specimen has just turned up with side = 1.511084, making it fairly certain that the last word remains to be spoken ...
Fred Lunnon
On 9/21/13, Fred Lunnon <fred.lunnon@gmail.com> wrote:
On 9/20/13, Dan Asimov <dasimov@earthlink.net> wrote:
Consider the unit n-cube and a largest regular n-simplex -- by volume -- that can fit inside it.
What is V(n), the n-dimensional volume of that simplex, as a function of n?
I think the first three of these are V(1) = 1, V(2) = 2sqrt(3) - 3, and V(3) = 1/3.
If this is too hard to calculate in general, maybe there is an asymptotic formula for V(n) ?
--Dan
Warren Smith: << If you go down up to 3 dimensions to get 3 mod 4, use Hadamard to inscribe regular simplex in cube-face (of the reduced dimension) then build back up, then you will not lose out too horribly (although this will not be optimal). >>
So a casual glance suggests that we might get an estimate for the answer in 4-space by fitting the base of a regular simplex of edge sqrt2 into one cube facet of the unit tesseract, giving [0, 0, 0, 0], [0, 0, 1, 1], [0, 1, 0, 1], [0, 1, 1, 0], [s, 1/2, 1/2, 1/2] where s = sqrt(5/4) .
However this doesn't apparently work very well: for one thing, the altitude s > 1 , so the simplex must be scaled down to edge sqrt(8/5) to fit inside the tesseract. For another, the configuration turns out to be a local maximum: any near-identity similarity which retains the simplex within the tesseract also shrinks it.
The maximal regular simplex turns out instead to have vertex coordinates [1, 0, 0, 0], [0, 1, 0, 0], [0, 0, 1, 0], [0, 0, 0, 1], [t/2, t/2, t/2, t/2] where t = (sqrt5 + 1)/2 is the "golden section" number tau, phi, etc. Which also has edge sqrt2 --- despite everything, Warren's estimate gave the correct answer exactly!
The corresponding content V(4) = rt5/24 ~ 0.093169 .
Fred Lunnon
Just in case anybody was wondering (evidently nobody was --- sniff!) the missing constant in my n = 6 construction is a = 1/2 (errm --- just testing you ...) Collecting results to date, the following tabulates (technically, lower bounds on) <dimension, side, content> for the maximal regular simplex inscribed in a unit hypercube: <0, 1.000000000000000, 1.000000000000000>, <1, 1.000000000000000, 1.000000000000000>, <2, 1.035276180410083, 0.4641016151377546>, <3, 1.414213562373095, 0.3333333333333333>, <4, 1.414213562373095, 0.09316949906249124>, <5, 1.511173070871356, 0.02843743486232121>, <6, 1.581138830084190, 0.007177059763087539>, <7, 2.000000000000000, 0.006349206349206349> ]; Another generalisation which occurs to me concerns the maximum content of an arbitrary simplex inscribed in the hypercube. Might the maximal simplex be regular? I don't know the answer, even for 2-space. Fred Lunnon On 9/25/13, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Hurtling unstoppably on whither angles fear to tread, a quick blunder around in 6-space led surprisingly quickly to critical (maximal?) side and content
sqrt(5/2) = 1.581138830084190, 0.007177059763087539
with vertices
[b,c,b,c,b,c], [1,a,0,1,1,1], [0,1,1,1,1,a], [1,1,1,a,0,1], [1,1,a,0,1,0], [a,0,1,0,1,1], [1,0,1,1,a,0]]
where
b = 0.08856217223385236, c = 0.3628540574112841
satisfy
b^2 - (3/2)b + 1/8 = 0, c = (b + 1)/3 .
There is 3-cyclic symmetry, and simplex vertices lie on hypercube edges, except for a single one in the interior.
As Warren pointed out, in 7-space we have the obviously maximal Hadamard simplex with side = 2 , sharing all vertices with the hypercube.
Fred Lunnon
On 9/23/13, Fred Lunnon <fred.lunnon@gmail.com> wrote:
At this juncture, the intelligent thing to do is abandon my gruesomely slowly-convergent search program, and instead solve for a,b,c maximising the side length of an equilateral simplex having this nice symmetric form. Lagrange finds only a single interior minimum; escaping to the nearby boundary by setting a = 0 , Lagrange locates a single maximum.
The upshot has
side = 1.511173070871355
with vertices
[0, 0, 0, 0, 0], [1, 1, 0, c, b], [1, 0, c, b, 1], [0, c, b, 1, 1], [c, b, 1, 1, 0], [b, 1, 1, 0, c]
where
b = 0.09303505049476144, c = 0.5243934872843100
are quartic irrationals such that b^4 - 6*b^3 + 14*b^2 - 12*b + 1 = 0 and c = b^3 - 4*b^2 + 6*b .
The 5-cyclic symmetry of the complete configuration is obvious above; all but one simplex vertex lie on hypercube (plane) faces.
This must almost certainly be the overall maximal possible regular simplex inscribable in a unit hypercube in 5-space.
Fred Lunnon
On 9/23/13, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Heady with success in 4-space, I embarked enthusiastically upon a general program for n-space --- only to recall, yet again, that it is better to travel hopefully than to arrive.
The 5-space case is truly horrible: there are evidently multiple local maxima all with side approximately 1.5, corresponding to simplex vertices lying on various combinations of boundary elements of the hypercube. While one simplex vertex always nestles securely in a hypercube vertex, just deciding whether other simplex vertices actually meet a hypercube line, plane etc. is in practice difficult because of ill-conditioning.
My most persuasive candidate so far in 20 attempts, by reason of symmetry, has also the second largest known side = 1.510898923478 ; its vertices are [1, 1, 1, a, b, c], [1, c, b, 1, a, 1], [1, a, c, b, 1, 1], [1, 1, a, c, 1, b], [1, b, 1, 1, c, a], [1, 0, 0, 0, 0, 0], where a := 0.003086685229, b := 0.089689781542, c := 0.524177233794; though these constants don't appear to be anything simply algebraic.
But alas for elegance --- a much gnarlier specimen has just turned up with side = 1.511084, making it fairly certain that the last word remains to be spoken ...
Fred Lunnon
On 9/21/13, Fred Lunnon <fred.lunnon@gmail.com> wrote:
On 9/20/13, Dan Asimov <dasimov@earthlink.net> wrote:
Consider the unit n-cube and a largest regular n-simplex -- by volume -- that can fit inside it.
What is V(n), the n-dimensional volume of that simplex, as a function of n?
I think the first three of these are V(1) = 1, V(2) = 2sqrt(3) - 3, and V(3) = 1/3.
If this is too hard to calculate in general, maybe there is an asymptotic formula for V(n) ?
--Dan
Warren Smith: << If you go down up to 3 dimensions to get 3 mod 4, use Hadamard to inscribe regular simplex in cube-face (of the reduced dimension) then build back up, then you will not lose out too horribly (although this will not be optimal). >>
So a casual glance suggests that we might get an estimate for the answer in 4-space by fitting the base of a regular simplex of edge sqrt2 into one cube facet of the unit tesseract, giving [0, 0, 0, 0], [0, 0, 1, 1], [0, 1, 0, 1], [0, 1, 1, 0], [s, 1/2, 1/2, 1/2] where s = sqrt(5/4) .
However this doesn't apparently work very well: for one thing, the altitude s > 1 , so the simplex must be scaled down to edge sqrt(8/5) to fit inside the tesseract. For another, the configuration turns out to be a local maximum: any near-identity similarity which retains the simplex within the tesseract also shrinks it.
The maximal regular simplex turns out instead to have vertex coordinates [1, 0, 0, 0], [0, 1, 0, 0], [0, 0, 1, 0], [0, 0, 0, 1], [t/2, t/2, t/2, t/2] where t = (sqrt5 + 1)/2 is the "golden section" number tau, phi, etc. Which also has edge sqrt2 --- despite everything, Warren's estimate gave the correct answer exactly!
The corresponding content V(4) = rt5/24 ~ 0.093169 .
Fred Lunnon
Yeah, I wondered the same thing. My guess is, It's true. But that ain't a proof. On 2013-09-25, at 12:58 PM, Fred Lunnon wrote:
Another generalisation which occurs to me concerns the maximum content of an arbitrary simplex inscribed in the hypercube. Might the maximal simplex be regular? I don't know the answer, even for 2-space.
Also, Fred, those are very nice maximal regular simplex results. Can you please say -- or repeat if I missed it -- how your optimizer program works? --Dan
On 9/25/13, Dan Asimov <dasimov@earthlink.net> wrote:
Yeah, I wondered the same thing. My guess is, It's true. But that ain't a proof.
On 2013-09-25, at 12:58 PM, Fred Lunnon wrote:
Another generalisation which occurs to me concerns the maximum content of an arbitrary simplex inscribed in the hypercube. Might the maximal simplex be regular? I don't know the answer, even for 2-space.
Also, Fred, those are very nice maximal regular simplex results. Can you please say -- or repeat if I missed it -- how your optimizer program works?
--Dan
No great imaginative leap to reveal, I'm afraid. The search program just plonks a little simplex at the centre of the hypercube, then repeatedly applies a random rotation, followed by translating to touch the hypercube boundary, and dilating to touch the opposite half --- assuming that the entire transformation increases the simplex diameter. The rotation angle is bounded; after a specified number of consecutive failures to improve, the angle bound is reduced by a specified ratio. Then I have to tinker with input values and trawl through the results, on the lookout for an exceptionally large side or (approximate) symmetry. If I get lucky, I replace the clusters of numerically close coordinate components by variables, then maximise the resulting side analytically, subject to the regularity constraint. So far, the answer has always come out satisfyingly just a little larger still than the largest experimental value. The Monte-Carlo search stage is of course excruciatingly slow to converge: partly on account of the high dimension (n+1)n/2 ; and partly because it is iterating towards a critical value, an inherently ill-conditioned procedure. Fred Lunnon
On 9/25/13, Dan Asimov <dasimov@earthlink.net> wrote:
On 2013-09-25, at 12:58 PM, Fred Lunnon wrote:
Another generalisation which occurs to me concerns the maximum content of an arbitrary simplex inscribed in the hypercube. Might the maximal simplex be regular? I don't know the answer, even for 2-space.
Yeah, I wondered the same thing. My guess is, It's true. But that ain't a proof.
Ho-ho --- equilateral has max area 0.4641 --- [[0, 0], [0, 1], [1, 0]] has area 0.5 ! WFL
Oops, now I know how that works in 2D. --Dan On 2013-09-25, at 6:24 PM, Fred Lunnon wrote:
On 9/25/13, Dan Asimov <dasimov@earthlink.net> wrote:
On 2013-09-25, at 12:58 PM, Fred Lunnon wrote:
Another generalisation which occurs to me concerns the maximum content of an arbitrary simplex inscribed in the hypercube. Might the maximal simplex be regular? I don't know the answer, even for 2-space.
Yeah, I wondered the same thing. My guess is, It's true. But that ain't a proof.
Ho-ho --- equilateral has max area 0.4641 --- [[0, 0], [0, 1], [1, 0]] has area 0.5 !
WFL
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A probably maximal 8-space regular simplex inscribed in the unit hypercube, with side = 2 and vertices [ a, 0, a, 0, a, 0, a, 0 ], [ 1, 1, 1, 1, 1, 1, 0, 0 ], [ 0, 1, 0, 1, 1, 0, 0, 1 ], [ 1, 1, 1, 1, 0, 0, 1, 1 ], [ 0, 1, 1, 0, 0, 1, 0, 1 ], [ 1, 1, 0, 0, 1, 1, 1, 1 ], [ 1, 0, 0, 1, 0, 1, 0, 1 ], [ 0, 0, 1, 1, 1, 1, 1, 1 ], [ 0, 1, 0, 1, 0, 1, 1, 0 ], where (as is customary) a := 1/2 ; symmetry is 4-cyclic. Two things I a find difficult to credit about this result: the first is that my spatchcock algorithm is still capable of producing any result at all; the second is that Warren has yet again opined what appears surely to be a sharply maximal result, on the grounds of what he admits was a goofball reason ... Updated <dim, side, content> table: <0, 1.000000000000000, 1.000000000000000>, <1, 1.000000000000000, 1.000000000000000>, <2, 1.035276180410083, 0.4641016151377546>, <3, 1.414213562373095, 0.3333333333333333>, <4, 1.414213562373095, 0.09316949906249124>, <5, 1.511173070871356, 0.02843743486232121>, <6, 1.581138830084190, 0.007177059763087539>, <7, 2.000000000000000, 0.006349206349206349>, <8, 2.000000000000000, 0.001190476190476190>; (technically, lower bounds on maxima achievable). Speechless! Fred Lunnon
Still contemplating Dan's n-space regular simplices inscribed in a unit hypercube, the way one does, for n = 9 now. Given that for n = 7,8 the (conjectural) max side equals 2 , it was not a promising start that my rotation-jiggling search program took 12 hours to find just a couple of examples exceeding that: but the good-ish news is that for n = 9 a lower bound on the maximal side is 2.00452263 . Such configurations as warrent further investigation do not seem to have a great deal evident in the way of symmetry. More promising is their preponderance --- in one case two-thirds --- of coordinate components near --- within say 0.02 --- of an integer --- 0 or 1 --- which, one may assume, are on the way to eventually converge to interval endpoints. The freedom 1 + (n+1)n/2 similarity transforming current into limiting configuration is therefore by this stage well over-determined. If only there were an effective algorithm to compute it ... I ran up program which forces these near-endpoints to endpoints, in the process, destroying the regularity of the simplex, then attempts to restore regularity by minimising the side-length variance, with respect to individual cooordinate components iteratively. This works a treat on smaller cases already solved, but disappointingly fails to progress on the problem to hand. A less direct approach --- solving for the almost orthogonal projective matrix of the final transformation --- involves large numbers of simultaneous quadratic equations. This looks fairly horrible --- "quadratic programming", perhaps? Fred Lunnon
I got a little bit lucky for n = 9 , guessing a configuration which plonks one simplex vertex just outside the hypercube --- but scaled down to fit, still gives lower bound 9*(2*Sqrt(5) - 3*Sqrt(2)) = 2.065457... on the maximal side length --- but that is unlikely to be sharp. The geometric problem I formulated earlier doesn't appear amenable to conventional "quadratic programming" or "convex programming" techniques, which apparently cope only with quadratic inequalities. It can be formulated in a nutshell: given the values of some subset comprising more than half the coordinate components of the vertices of a regular simplex, complete its coordinates. Fred Lunnon On 9/30/13, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Still contemplating Dan's n-space regular simplices inscribed in a unit hypercube, the way one does, for n = 9 now.
Given that for n = 7,8 the (conjectural) max side equals 2 , it was not a promising start that my rotation-jiggling search program took 12 hours to find just a couple of examples exceeding that: but the good-ish news is that for n = 9 a lower bound on the maximal side is 2.00452263 .
Such configurations as warrent further investigation do not seem to have a great deal evident in the way of symmetry. More promising is their preponderance --- in one case two-thirds --- of coordinate components near --- within say 0.02 --- of an integer --- 0 or 1 --- which, one may assume, are on the way to eventually converge to interval endpoints.
The freedom 1 + (n+1)n/2 similarity transforming current into limiting configuration is therefore by this stage well over-determined. If only there were an effective algorithm to compute it ...
I ran up program which forces these near-endpoints to endpoints, in the process, destroying the regularity of the simplex, then attempts to restore regularity by minimising the side-length variance, with respect to individual cooordinate components iteratively. This works a treat on smaller cases already solved, but disappointingly fails to progress on the problem to hand.
A less direct approach --- solving for the almost orthogonal projective matrix of the final transformation --- involves large numbers of simultaneous quadratic equations. This looks fairly horrible --- "quadratic programming", perhaps?
Fred Lunnon
My previous tentative prospect for the maximal regular simplex inscribed within a unit hypercube in 6-space with side sqrt(5/2) = 1.581138830084 ... has been toppled by a tantalising new candidate with side sqrt(11 - 6 sqrt(2)) = 1.585786437627 ... and vertices [ d, d, d, d, d, d ], [ b, 0, c, 0, 1, 1 ], [ 1, b, 0, c, 0, 1 ], [ 1, 1, b, 0, c, 0 ], [ 0, 1, 1, b, 0, c ], [ c, 0, 1, 1, b, 0 ], [ 0, c, 0, 1, 1, b ]; where d = 0.0257642624106 , c = 0.7071067811865475 , b = 0.4142135623730950 satisfy b = sqrt2 - 1, c = sqrt/2, d^2 - (c + 1/3)d + (c 4/3 - 11/12) = 0 . Symmetry is 6-cyclic; the fixed vertex lies diagonally from the origin, separated from it by merely 1 part in 40 . One might reasonably expect this to be outperformed by some 6-cyclic contender having one vertex squarely at the origin. But it seems one would be mistaken: an exhaustive search found just three such candidates, with approximate sides 1.498028 , 1.511416 , 1.551629 . These constructions give only lower bounds on the maximum side obtainable. A straightforward upper bound sqrt(n+1)/2 = sqrt(7/2) = 1.870828693387 comes via the circumradius of the hypercube dominating that of the inscribed simplex. Less trivially, Charles Greathouse's Hadamard observation leads to slightly improved upper bound ( (1/80) / (sqrt(7/128)/720) )^(1/6) = 1.837453732988 . Neither is likely to be anywhere near as close as the current lower bound, which I expect be sharp to at least 2 decimals. Fred Lunnon On 10/1/13, Fred Lunnon <fred.lunnon@gmail.com> wrote:
I got a little bit lucky for n = 9 , guessing a configuration which plonks one simplex vertex just outside the hypercube --- but scaled down to fit, still gives lower bound 9*(2*Sqrt(5) - 3*Sqrt(2)) = 2.065457... on the maximal side length --- but that is unlikely to be sharp.
The geometric problem I formulated earlier doesn't appear amenable to conventional "quadratic programming" or "convex programming" techniques, which apparently cope only with quadratic inequalities. It can be formulated in a nutshell: given the values of some subset comprising more than half the coordinate components of the vertices of a regular simplex, complete its coordinates.
Fred Lunnon
On 9/30/13, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Still contemplating Dan's n-space regular simplices inscribed in a unit hypercube, the way one does, for n = 9 now.
Given that for n = 7,8 the (conjectural) max side equals 2 , it was not a promising start that my rotation-jiggling search program took 12 hours to find just a couple of examples exceeding that: but the good-ish news is that for n = 9 a lower bound on the maximal side is 2.00452263 .
Such configurations as warrent further investigation do not seem to have a great deal evident in the way of symmetry. More promising is their preponderance --- in one case two-thirds --- of coordinate components near --- within say 0.02 --- of an integer --- 0 or 1 --- which, one may assume, are on the way to eventually converge to interval endpoints.
The freedom 1 + (n+1)n/2 similarity transforming current into limiting configuration is therefore by this stage well over-determined. If only there were an effective algorithm to compute it ...
I ran up program which forces these near-endpoints to endpoints, in the process, destroying the regularity of the simplex, then attempts to restore regularity by minimising the side-length variance, with respect to individual cooordinate components iteratively. This works a treat on smaller cases already solved, but disappointingly fails to progress on the problem to hand.
A less direct approach --- solving for the almost orthogonal projective matrix of the final transformation --- involves large numbers of simultaneous quadratic equations. This looks fairly horrible --- "quadratic programming", perhaps?
Fred Lunnon
participants (5)
-
Dan Asimov -
Eugene Salamin -
Fred Lunnon -
Hans Havermann -
Warren D Smith